Question 15.81


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104822659
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Question 15.81

Postby 104822659 » Sun Feb 19, 2017 9:25 pm

Hi, I'm not understanding the loss and reverse part of the question. If anyone can help, that would be great!
The equilibrium constant for the second-order attachment of a substrate to the active site of an enzyme was found to be 326 at 310 K. At the same temperature, the rate constant for the second-order attachment is 7.4 x10^ 7 L*mol^-1*s^-1. What is the rate constant for the loss of unreacted substrate from the active site (the reverse of the attachment reaction)?

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Re: Question 15.81

Postby Chem_Mod » Mon Feb 20, 2017 12:19 am

The question is really just asking the relationship between forward reaction rate constant k, reverse reaction rate constant k' and the equilibrium constant K.


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